Using the housework data where employed men spend an average mu = 8.2 hours/week with sigma = 2.1 hours (normal distribution), P(X geq 9) is found by computing z as:

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CSS Statistics QUESTION #2004

Question 1

Using the housework data where employed men spend an average \(\mu = 8.2\) hours/week with \(\sigma = 2.1\) hours (normal distribution), \(P(X \geq 9)\) is found by computing \(z\) as:

\(z = 0.286\)
\(z = 0.381\)✔️
\(z = 0.476\)
\(z = 0.524\)

Correct Answer Explanation

\(z = \dfrac{9 - 8.2}{2.1} = \dfrac{0.8}{2.1} \approx 0.381\). Then \(P(X \geq 9) = P(Z \geq 0.381) = 1 - \Phi(0.381) \approx 1 - 0.6484 = 0.3516\).

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